Question 1 - Constructing a triangle when 2 sides and angle between is known (SAS) - Practical Geometry

Last updated at April 16, 2024 by

Ex 10.3, 1 - Construct DEF such that DE = 5 cm, DF = 3 cm, EDF = 90

Ex 10.3, 1 - Chapter 10 Class 7 Practical Geometry - Part 2
Ex 10.3, 1 - Chapter 10 Class 7 Practical Geometry - Part 3
Ex 10.3, 1 - Chapter 10 Class 7 Practical Geometry - Part 4
Ex 10.3, 1 - Chapter 10 Class 7 Practical Geometry - Part 5

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Question 1 Construct ∆DEF such that DE = 5 cm, DF = 3 cm and m∠EDF = 90°. First we draw a rough sketch To construct it, We follow these steps Steps of construction 1. Draw a line segment DF of length 3 cm 2. Now, we draw 90° from point D Check Ex 11.1, 1 of Chapter 11 Class 9 NCERT to see how to draw 90° 3. Taking D as center, 5 cm as radius, we draw an arc Let the point where arc intersects the ray be point E 4. Join EF and label the sides Thus, Δ DEF is the required triangle

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo